What is expand ((x-6)(x-9))/(10) ?

The solution to expand ((x-6)(x-9))/(10) is (x^2}{10}-\frac{3x)/2+27/5

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expand ((x-6)(x-9))/(10)

expand

\frac{( x - 6 ) ( x - 9 )}{10}

\frac{x ^{2}}{10} - \frac{3 x}{2} + \frac{27}{5}

Solution steps

\frac{( x - 6 ) ( x - 9 )}{10}

Expand

(x - 6) (x - 9) : x^{2} - 15 x + 54

= \frac{x ^{2} - 15 x + 54}{10}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{x ^{2} - 15 x + 54}{10} = \frac{x ^{2}}{10} - \frac{15 x}{10} + \frac{54}{10}

= \frac{x ^{2}}{10} - \frac{15 x}{10} + \frac{54}{10}

Cancel

\frac{15 x}{10} : \frac{3 x}{2}

= \frac{x ^{2}}{10} - \frac{3 x}{2} + \frac{54}{10}

Cancel

\frac{54}{10} : \frac{27}{5}

= \frac{x ^{2}}{10} - \frac{3 x}{2} + \frac{27}{5}

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\frac{( - x + 14 ) \cdot ( x - 6 )}{2}

\frac{( 3 x )}{( x + 8 ) ^{2}}

\frac{x ( x ^{2} - x ) - 4 ( x + 1 )}{x - 1}

What is expand ((x-6)(x-9))/(10) ?
The solution to expand ((x-6)(x-9))/(10) is (x^2}{10}-\frac{3x)/2+27/5